Math Help Forum: Noraml distribution question...

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    Question Noraml distribution question...

    Supppose the life span of a battery is normally distributed. The useful life of the battery has a mean of 600 hours and standard deviation of 40 hours. Find the probability that a battery chosen at random will last between 600 and 680 hours. I need a little help getting going on this one. I know it is somewhere between .45 and .55, but am unsure how to get there. Any help is greatly appreciated.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by bemidjibasser View Post
    Supppose the life span of a battery is normally distributed. The useful life of the battery has a mean of 600 hours and standard deviation of 40 hours. Find the probability that a battery chosen at random will last between 600 and 680 hours. I need a little help getting going on this one. I know it is somewhere between .45 and .55, but am unsure how to get there. Any help is greatly appreciated.
    Let X represent the life span of a battery. \mu=600 and \sigma=40

    We are looking for P\!\left(600<X<680\right)

    Incorporating the z-score, we see that P\!\left(600<X<680\right)=P\!\left(600-600<x-\mu<680-600\right) =P\!\left(0<\frac{x-\mu}{\sigma}<2\right)=P\!\left(0<Z<2\right).

    You can find this solution in standard normal distribution tables. If you have a TI 83 or higher, you can use the normalcdf command to give you your probability. It should be about .4772.
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